Particles and Waves:
We have so far discussed two behaviors of light: straight line motion (Geometric Optics) and the wave-like behavior (Wave Optics). In this chapter, the particle-like behavior of light will be discussed. In fact, the particle-like behavior is also associated with a frequency and it cannot be separated form the wave-like behavior (Quantum Optics).
Max Planck formulated this theory that as electrons orbit the nucleus of an atom, they receive energy from the surroundings in different forms. Typical forms of receiving energy are: heat waves, light waves, and collision with other electrons and particles. The radius at which an electron orbits is a function of electron's energy and therefore electron's speed as well as its distance from the nucleus. Recall that K.E. = (1/2)Mv2. Each electron is also under a Coulomb attraction force from the nucleus given by F = ke2/r2. Furthermore, circular motion requires a centripetal force Fc = Mv2/r. We know that it is the Coulomb force F that provides the necessary centripetal force Fc for the electron rotation about the nucleus.
The above discussion clarifies that, in simplest explanation, each electron takes a certain radius of rotation depending on its energy or speed. When an electron receives extra energy, it then has to change its orbit or radius of rotation. It has to take an orbit of greater radius. The radius it takes is not just any radius. The radii of rotation for any electron orbiting a nucleus are discrete values that will be discussed in the chapter on atomic physics. When such transition occurs, a vacant orbit is left behind that must be filled. It may be filled by the same electron or any other one. The electron that fills that vacant orbit must have the correct energy that matches the energy of that orbit. The electron that fills that orbit may have excess energy that has to be given off before being able to fill that vacant orbit. The excess energy that an electron gives off appears as a burst of energy, a parcel of energy, a packet of energy that is called a "quantum of energy" according to Max Planck.
The excess energy is simply the energy difference between two different orbits. If an electron returns from a greater radius orbit Rm with an energy level Em to a smaller radius orbit Rn with an energy level En, it releases a quantum of energy equal to the energy difference Em- En. Max Planck showed that this energy difference is proportional to the frequency f of that particular transition. The proportionality constant is h with a value of h = 6.626x10-34 J.s called the " Planck's constant." The packet or quantum of energy is also called a "photon."
In electron-volts, h has a value of h = 4.14x10-15eV-sec. The Plancks' formula is:
Em - En = hf
|or, ΔE = hf||or, E = hf .|
Example 1: Calculate (a) the energy of photons with a frequency of 3.2x1014 Hz. (b) Find their corresponding wavelength and (c) express if they are in the visible range.
Solution: (a) ΔE = hf ; ΔE = ( 6.626x10-34 J.s )( 3.2x1014 /s) = 2.12x10-19 J
Note that 1eV = 1.6x10 -19J. Our answer is a little more than 1eV. In fact it is (2.12 /1.6) = 1.3 eV.
(b) c = f λ ; λ = (3.00x108m/s)/(3.2x1014/s) = 940x10-9m = 940nm.
(c) The visible range is between 400 nm - 700 nm. This photon is not in the visible range. It is infrared.
Example 2: Calculate (a) the energy (in Joules) of each photon of ultraviolet light with a wavelength of 225nm. (b) Convert that energy to electron-volts.
Solution: (a) E = hf and f = c/λ ; therefore, E = hc/λ.
E = 6.626x10-34J.s (3.00x108m/s)/225x10-9m = 8.83x10-19 J.
(b) Since 1eV = 1.6x10-19J ; therefore, ΔE = 5.5eV.
The Photoelectric Effect:
The mechanism by which photoelectric effect operates may be used to verify the particle-like behavior of light. A photoelectric cell can be made of a vacuum tube in which two metallic plates or poles are fixed. The two plates are connected to two wires that come out of the sealed glass tube and are used for connection to other electronic components. For time being, let us connect a photoelectric cell to just a galvanometer (sensitive ammeter) as shown in the figure below. One terminal (plate) in the tube may be mounted in a slanted fashion in order for the light coming from outside to effectively shine on it. This side forms the negative pole. The other side collects or receives electrons and forms the positive pole.
| When photons of light are sent
toward the metal plate, it is observed that the galvanometer in the
circuit shows the passage of a current. When the light is
cut off, the current stops. This shows that the collision of
photons of light on the metal surface must release electrons from
the outer shells of the outermost atomic layers of the metallic
Each energetic photon that collides with the metal surface, releases one electron. This released electron has some speed and therefore some K.E. = 1/2Mv2. The atoms of the outer surface that have lost electrons, replenish their electron deficiencies from the inner layer atoms of the metal oxide or the nearby free electrons.
This replenishing process transmits layer by layer through the wire and the galvanometer all the way to the pole labeled "Positive."
The positive end pulls the released electrons from the negative end through the vacuum tube and the circuit completes itself. This process occurs very fast. As soon as light hits the metal plate, the circuit is on. As soon as light is cut off, the circuit goes off.
The conclusion of the above experiment is that photons of light act as particles and kick electrons out of their orbit. This verifies the particle-like behavior of light.
The Photoelectric Cell Formula:
The energy necessary to just detach an electron out of a metal surface is called the " Work Function" of that metal and is shown by Wo . If the energy of each incident photon on the metal surface is hf, and the kinetic energy of the released electron is K.E., then we may write the following energy balance for a photoelectric cell:
hf = Wo + K.E.
According to this equation, hf must be greater than Wo for an electron to be released. Since h is the Planck's constant; therefore, f must be high enough for the photon to be effective. There is a limit for frequency below which nothing happens. That limit happens when the frequency of the incident photon is just enough to release an electron. Such released electron has a K.E. = 0. At the limiting frequency called the "threshold frequency", the K.E. of the released electron is zero. With K.E. = 0 and replacing f by fth, we get:
h fth = Wo
|or||fth = Wo / h.|
The above formula gives the threshold frequency, fth .
Example 3: The work function of the metal plate in a photoelectric cell is 1.73eV. The wavelength of the incident photons is 366nm. Find (a) the frequency of the photons, (b) the K.E. of the released electrons, and (c) the threshold frequency and wavelength for this photoelectric cell.
(a) c = f λ ; f = c/λ= (3.00x108m/s)/(366x10-9m) = 8.20x1014 Hz.
(b) hf = W0 + K.E. ; K.E. = hf - W0
K.E. = (4.14x10-15eV∙s) (8.20x1014 s-1) - 1.73eV = 1.66eV.
(c) fth = Wo/h ; fth = 1.73eV/(4.14x10-15eV∙s) = 4.18x1014 Hz.
λth = c/fth ; λth = (3.00x108m/s)/(4.18x1014 Hz) = 718nm.
According to de Broglie, for every moving particle of momentum Mv, we may associate an equivalent wavelength λ describing its wave motion behavior such that
de Broglie Wavelength:
where λ is called the "de Broglie wavelength."
Example 4: Calculate the DeBroglie wavelength associated with the motion of an electron that orbits a hydrogen atom at a speed of 6.56x106 m/s.
Solution: Using λ = h/Mv, we may write:
λ = 6.626x10-34J∙s/[(9.108x10-31kg)(6.56x106 m/s)] = 1.11x10-10m.
Chapter 29 Test Yourself 1:
1) The energy of a photon of light, according to Max Planck's formula is (a) E = 1/2Mv2 (b) E = hf (c) E = Mgh.
2) The Planck's is (a) 6.6262x10-34 J∙s (b) 4.14x10-15 eV∙s (c) both a & b. click here.
3) An electron orbiting the nucleus of an atom can be energized by (a) receiving a heat wave (b) getting collided by another subatomic particle (c) by getting hit by a photon (d) both a, b, & c.
4) When an electron is energized by any means, it requires (a) a greater radius of rotation (b) a smaller radius of rotation (c) it stays in the same orbit but spins faster. click here.
5) When there is a vacant orbit, it will be filled with an electron from (a) a lower orbit (b) a higher orbit.
6) A higher orbit means (a) a greater radius (b) a faster moving electron (c) a greater energy (d) a, b, and c.
7) The excess energy that an electron in a higher orbit has is released in the form of a photon (a small packet or burst of energy) as the electron fills up a lower orbit. (a) True (b) False click here.
8) The excess energy is (a) the energy difference, E2 - E1, of the higher and lower orbits (b) the energy each electron has anyway (c) both a & b.
9) A photon has a mass of (a) zero (b) 1/2 of the mass of an electron (c) neither a nor b.
10) Each photon carries a certain amount of energy. We may use the Einstein formula (E = Mc2) and calculate an equivalent mass for a photon. (a) True (b) False click here.
11) The greater the energy of a photon (a) the higher its speed (b) the higher its velocity (c) the higher it frequency (d) a, b, c, & d.
12) The greater the energy of a photon the lower its wavelength. (a) True (b) False
13) The formula for waves speed, v = f λ, takes the form of (a) c = f λ for photons of visible light only (b) for photons of non-visible light only (c) for the full spectrum of E&M waves that visible light is a part of. click here.
Problem: A student has calculated a frequency of 4.8x1016 Hz for a certain type of X-ray and a wavelength of 7.0nm.
14) Use the equation v = f λ and calculate v to see if the student's calculations is correct. (a) Correct (b) Wrong
15) The answer to Question 14 is (a) 3.36x108 m/s (b) 3.36x1017m/s (c) neither a nor b. click here.
16) The reason why the answer to Question 14 is wrong is that v turns out to be greater than the speed of light in vacuum that is 3.0x108 m/s. (a) True (b) False
17) In the photoelectric effect, (a) electrons collide and release photons (b) photons collide and release electrons (c) neither a nor b. click here.
18) In a photoelectric cell, the plate that receives photons, becomes (a) negative (b) positive (c) neutral.
19) The reason why the released (energized) electrons do no return back to their shells is that (a) their energies are more than enough for the orbits they were in (b) the orbits (of the atoms of the metal plate) that have lost electrons, quickly replenish electrons from the inner layer atoms of the metal plate (c) the outer shells atoms that have lost electrons will be left in loss for ever (d) a & b. click here.
20) When light is incident on the metal plate of a photoelectric cell, the other plate or pole of the cell becomes positive. The reason is that (a) photons carry negative charges (b) the other pole loses electrons to replenish the lost electrons of the metal plate through the outside wire that connects it to the metal plate (c) both a & b.
21) In a photoelectric cell, the released electrons (a) vanish in the vacuum of the cell (b) accelerate toward the other pole because of the other pole is positive (c) neither a nor b.
22) The negative current in the external wire of a photoelectric cell is (a) zero (b) out of the metal plate (c) toward the negative plate. click here.
23) In a photoelectric cell, the energy of each incident photon is (a) 1/2Mv2 (b) hf (c) Wo.
24) In a photoelectric cell, the work function of the metal plate is named (a) 1/2Mv2 (b) hf (c) Wo.
25) In a photoelectric cell, the energy of each released electron is (a) 1/2Mv2 (b) hf (c) Wo. click here.
26) A 5.00eV incident photon has a frequency of (a) 1.21x10-15Hz (b) 1.21x1015Hz (c) 2.21x1015Hz.
27) A UV photon of f = 3.44x1015Hz has an energy of (a)14.2eV (b) 2.27x10-18J (c) a & b.
28) When 3.7eV photons are incident on a 1.7eV work function metal, each released electron has an energy of (a) 2.0eV (b) 5.4eV (c) 6.3eV. click here.
29) 4.7eV photons are incident on a 1.7eV work function metal. Each released electron has an energy of (a) 4.8x10-19J (b) 3.0eV (c) both a & b.
30) 3.7eV photons incident on a 1.7eV work function metal cause released electrons of speed (a) 8.4x10-5m/s (b) 8.4x10 5m/s (c) 8.4x10-15m/s. click here.
31) A speed of 8.4x10-5m/s is not reasonable for a moving electron because (a) electrons always move at the speed of light (b) this speed has a power of -5 that makes it very close to zero same as being stopped (c) neither a nor b.
32) If the released electrons in a photoelectric effect have an average speed of 9.0x105 m/s and the energy of the incident photons on the average is 4.0eV, the work function of the metal is (a) 1.3eV (b) 1.1eV (c) 1.7eV. click here.
33) The wavelength associated with the motion of proton at a speed of 6.2x106m/s is(a) 6.4x10-14m (b) 9.4x10-14m (c) 4.9x10-14m.
34) The diameter of hydrogen atom (the whereabouts of its electronic cloud) is 0.1nm or 10-10m called "Angstrom." The diameter of the nucleus of the hydrogen atom is even 100,000 times smaller or10-15m called "Femto-meter (fm)." The wavelength associated with the moving proton in Question 33 is (a) 6.4fm (b) 64fm (c) 640fm. click here.
1) Calculate (a) the energy of photons with a frequency of 6.40x1014Hz. (b) Find their corresponding wavelength and (c) express if they are in the visible range.
2) Calculate (a) the energy (in Joules) of each photon of UV light with a wavelength of 107nm. (b) Convert that energy to electron-volts.
3) The work function of the metal plate in a photoelectric cell is 2.07eV. The wavelength of the incident photons on it is 236nm. Find (a) the frequency of the photons, (b) the energy of each, (c) the K.E. of the released electrons, (d) their speed, and (e) the threshold frequency and wavelength for this photoelectric cell.
4) Calculate the De Broglie wavelength associated with the motion of an electron that hast a speed of (a) 1.31x107m/s.
1) 2.65eV, 469nm, Yes 2) 1.86x10-18J, 12eV
3) 1.27x1015Hz, 5.26eV, 3.19eV, 1.1x106m/s, 5.00x1014Hz, 600.nm
4) 0.556 Angstrom.